Continous Functions

Convergence

Definition. [Limit of a Function] A function $f$ tends to a limit $l$ as $x \to c$ if

\[(\forall \varepsilon > 0)(\exists \delta > 0)(\forall x : 0 < |x - c| < \delta)\;|f(x) - l| < \varepsilon\]

We then also write

\[\lim_{x \to c} f(x) = l\]

The value of $\delta$ depends on $\varepsilon$ and can be emphasized by writing $\delta(\varepsilon)$. Also, the value of $f(c)$ is not necessarily defined for the limit to exist.

The convergence of function can be characterized in terms of convergent sequences.

Proposition. $f(x) \to l$ as $x \to c$ iff for all sequences $(a_n)$ in which $a_n \to c$ then $f(a_n) \to l$ as $n \to \infty$.

Proof.

($\Rightarrow$) Suppose $f(x) \to l$ as $x \to c$ and $(a_n)$ is a sequence in which $a_n \to c$. Then for all $\varepsilon > 0$, there exists a $\delta > 0$ such that $\vert f(x) - l \vert < \varepsilon$ when $0 < \vert x - c \vert < \delta$. As $a_n \to c$, there exists an $N$ such that for all $n > N$, $\vert a_n - c \vert < \delta$ and therefore $\vert f(a_n) - l \vert < \varepsilon$.

($\Leftarrow$) Contrapositive. Suppose $f(x)$ does not converge to $l$ as $x \to c$. Then there exists $\varepsilon > 0$ such that we find no suitable $\delta > 0$. For each $n \in \mathbb{N}$, there exists $x_n$ such that $\vert x_n - 1/n \vert < 0$ with $\vert f(x_n) - l \vert \ge \varepsilon$. Therefore, we can always find a sequence in which $x_n \to c$ as $n \to \infty$ but $f(x_n)$ does not converge to $l$.

We can conclude the general principle of convergence for functions similar to that for sequences.

Theorem. The following statements are equivalent.

  • There exists a $l$ such that $f(x) \to l$ as $x \to c$.

  • Whenever $(a_n)$ is a sequence which tends to $c$ then $(f(a_n))$ is a Cauchy sequence.

  • Given $\varepsilon > 0$, there exists $\delta > 0$ such that if $\vert x - c \vert < \delta$ and $\vert y - c \vert < \delta$ then $\vert f(x) - f(y) \vert < \varepsilon$.

Different from sequences, we can have one-sided convergence.

Definition. A function $f$ tends to a limit $l$ as $x$ tends to $c$ from the right if

\[(\forall \varepsilon > 0)(\exists \delta > 0)(\forall x : c < x < c + \delta)\;|f(x) - l| < \varepsilon\]

and denoted by

\[\lim_{x \to c+} f(x) = l\]

Similarily we can define that $x$ tends to $c$ from the left, i.e.

\[\lim_{x \to c-} f(x) = l\]

The limit of function exists iff $\lim_{x \to c+} f(x) = \lim_{x \to c-} f(x) = l$.

Continuity

Definition. A function $f$ is continuous at $c$ if $f(x) \to f(c)$ as $x \to c$.

Equivalently, we can say $f$ is continuous if

\[(\forall \varepsilon > 0)(\exists \delta > 0)(\forall x : |x - c| < \delta)\;|f(x) - f(c)| < \varepsilon\]

We can generalize the part $\vert x - c \vert < \delta$ as $c - \delta_1 < x < c + \delta_2$, where $\delta_1$ and $\delta_2$ can be different, and have any alternative definition of continuity.

Definition. An open interval is called a neighbourhood of any one of its points.

Definition. A function $f$ is continuous at $c$ if given any neighbourhood $N_1$ of $f(c)$, there is a neighbourhood $N_2$ of $c$ such that if $x \in N_2$ then $f(x) \in N_1$.

We can then extend the definition of continuity in an interval.

Definition. A function $f$ is continuous in the closed interval $[a, b]$ if

  • for each $c$ in $a < c < b$, $f$ is continuous at $c$;

  • $\lim_{x \to a+} f(x) = f(a)$ and $\lim_{x \to b-} f(x) = f(b)$.

A function $f$ is continuous in the open interval if it is continuous at each point of the interval.

Proposition. Similar to convergence of sequences,

  • sum of two continuous functions is continuous;

  • product of two continuous functions is continuous;

  • quotient of two continuous functions is continuous wherever the denominator is not zero.

Base on that we can conclude any polynomials and quotient of two polynomials (except for points where denominator is zero) are continuous.

Proposition. Suppose that $g(x)$ is continuous for $x = \xi$ and $g(\xi) = \eta$ and $f(y)$ is continuous at $y = \eta$. Then $f(g(x))$ is continuous at $x = \xi$.

Proof.

For all $\varepsilon > 0$, there exists $\delta > 0$ such that when $\vert y - \eta \vert < \delta$, $\vert f(y) - f(\eta) \vert < \varepsilon$. Also, there exists $\tau > 0$ such that when $\vert x - \xi \vert < \tau$, $\vert g(x) - g(\xi) \vert < \delta$. Combining the two we can conclude $f(g(x))$ is continuous at $x = \xi$.

Although composite of continuous functions is continuous, the limit of composite of functions might differ. For example, consider

\[f(x) = 0 \quad \text{and} \quad g(y) = \begin{cases} 1 & y \not= 0 \\ 0 & y = 0 \end{cases}\]

We have $f(x) \to 0$ as $x \to 0$ and $g(y) \to 1$ as $y \to 0$ but $g(f(x)) = g(0) = 0$ for all $x$ so $g(f(x)) \to 0$ as $x \to 0$. The statement is only true when $g(y)$ is continuous or $f(x) \not l$ when $x$ is in the neighbourhood of $a$.

Intermediate Value Theorem

Theorem. [Intermediate Value Theorem] Suppose that $f$ is continuous in the closed interval $[a, b]$ and $\eta$ is a number such that $f(a) < \eta < f(b)$. Then there exists $a < \xi < b$ such that $f(\xi) = \eta$. A similar result holds if $f(a) > f(b)$.

Proof.

Let $S = \Set{a \le x \le b : f(x) < \eta}$. $a \in S$ so $S$ is non-empty, and bounded above by $b$. Thus, the set $S$ has a supremum $\xi$.

Since $f$ is continuous at $a$, there is an interval $a \le x \le a + \delta$ such that $f(x) < \eta$ so $\xi \ge a + \delta > a$. Similarily there is an interval $b - \delta \le x \le b$ such that $f(x) > \eta$ so $\xi \le b - \delta < b$. Therefore, $a < \xi < b$ and $f$ is continuous at $\xi$.

If $f(\xi) > \eta$, say $f(\xi) = \eta + \varepsilon$. By continuity, there exists $\delta$ such that for $\xi - \delta < x < \xi$, $\vert f(\xi) - f(x) \vert < \varepsilon$. As $\xi$ is the supremum of $S$, there exists $x’ \in S$ such that $\xi - \delta < x’ < \xi$. Thus, $f(\xi) - f(x’) < \varepsilon$ which implies $f(x’) > \eta$ which is a contradiction. If $f(\xi) < \eta$, say $f(\xi) = \eta - \varepsilon$. By continuity, there exists $\delta$ such that for $\xi < x < \xi + \delta$, $f(x) - f(\xi) < \varepsilon$ and such $x$ is in $S$ which contradicts with $\xi$ being supremum. Hence, by trichotomy, $f\xi) = \eta$.

Alternatively, we can use the technique of repeated bisection to prove that. Let $a_0 = a$ and $b_0 = b$ and $m_0 = (a_0 + b_0) / 2$. If $f(m_0) < \eta$, we set $a_1 = m_0$ and $b_1 = b_0$, otherwise we set $a_1 = a_0$ and $b_1 = m_0$. Repeat this process recursively and the sequences $(a_n)$ is increasing and $(b_n)$ is decreasing. As $f(a_n) \le \eta \le f(b_n)$ and both $(a_n)$ and $(b_n)$ converges to a common limit $\xi$, we have $f(\xi) = \eta$.

Extreme Value Theorem

Theorem. [Extreme Value Theorem] If $f$ is continuous in the closed interval $[a, b]$, then $f$ is bounded and attains its bounds, i.e. there exists $y, z$ such that $f(y) = \sup f$ and $f(z) = \inf f$.

Proof.

Similar to the proof of Intermediate Value Theorem, we can use the supremum of a suitably defined set of values $x \in [a, b]$. Let $S$ be the set of numbers $x_1$ such that $f(x)$ is bounded above for $a \le x \le x_1$. Since $a \in S$, $S$ is non-empty and bounded above by $b$ and therefore has a supremum $\xi$. There are three possibilities (i) $a < \xi < b$, (ii) $\xi = a$, (iii) $\xi = b$. For (i), as $f$ is continuous at $\xi$, we can find an interval $(\xi - \delta, \xi + \delta)$ such that $f(x) < f(\xi) + 1$. It means $f(x)$ has to be bounded above for $a \le x \le \xi + \delta/2$ which contradicts to $\xi$ being the supremum. Similar arguments show that (ii) also leads to contradiction so $\xi = b$.

Bisect the interval $[a, b]$ such that $\sup f = M$ for $a_n \le x \le b_n$. We again have two sequences $(a_n)$ and $(b_n)$ which converges to the same limit $\xi$. Assume $f(\xi) = k < M$, since $f$ is continuous at $\xi$, we can always find an interval $[a_n, b_n]$ containing $\xi$ such that $\sup f(x) = M$ which is a contradiction. Hence, $f(\xi) = M$.

Alternatively, if $f$ is not bounded above, then for each $n$, we can find $x_n \in [a, b]$ such that $f(x_n) \ge n$ for all $n$. By Bolzano-Weierstrass, since $(x_n)$ is bounded, the sequence has a convergent subsequence $(x_{n_k})$ with limit $\xi$. Since $f$ is continuous at $\xi$, $f(x_{n_k}) \to f(\xi)$ which is a contradiction.

Let $M = \sup f$. Then for each $n$, we can find $x_n$ such that $f(x_n) \ge M - 1/n$. By Bolzano-Weierstrass, $(x_n)$ has a convergent subsequence $(x_{n_k})$ with limit $\xi$ and $M - 1/n_k \le f(x_{n_k}) \le M$. Hence, $f(\xi) = M$.

Inverse Functions

We can now give the conditions for the existence of inverse functions.

Proposition. Let $f$ be continuous and strictly increasing for closed interval $[a, b]$ and $f(a) = c$ and $f(b) = d$. Then there is a function $g$, continuous and strictly increasing for $[c, d]$, such that $f(g(x)) = x$.

Proof.

Let $k$ be any number such that $c < k < d$. By Intermediate Value Theorem, there is a $h$ such that $f(h) = k$. Since, $f$ is strictly increasing, $h$ is unique. Therefore, $f$ is both injective and surjective and hence bijective.

$g$ is strictly increasing since $f$ is strictly increasing. Also, suppose $g(k) = h$. For any $\varepsilon > 0$, $h - \varepsilon < g(y) < h + \varepsilon$ for $f(h - \varepsilon) < y < f(h + \varepsilon)$ and $g$ is continuous.

Uniform Continuity

Definition. Suppose that $f$ is a bounded real-valued function on a non-empty set $S$. The leap/oscillation of $f$ on $S$ is defined by

\[\Omega(f, S) = \sup \Set{|f(s) - f(t)| : s, t \in S} = \sup_{s \in S} f(s) - \inf_{s \in S} f(s)\]

Definition. A function $f$ is uniformly continuous in an interval $I$ if

\[(\forall \varepsilon > 0)(\exists \delta > 0)(\forall x_1, x_2 \in I : |x_1 - x_2| < \delta)\,|f(x_1) - f(x_2)| < \varepsilon\]

The difference to continuity is that instead of having a $\delta$ that works for a particular value of $x$, it works for every $x$ in the interval.

Lemma. Suppose that $f$ is continuous in the closed interval $[a, b]$. Then, given $\varepsilon$, the interval can be divided into a finite number of parts in each of which the leap of $f$ is less than $\varepsilon$.

Proof.

Suppose that the interval cannot be divided as described in the proposition. Bisect the interval $[a, b]$ and choose the half for which it is false (choose the left if both false) and denote it as $[a_1, b_1]$. Repeat this process and we have an increasing sequence $(a_n)$ and decreasing sequence $(b_n)$ with a common limit $\xi$.

Since $f$ is continuous at $\xi$, there is an interval $(\xi - \delta, \xi + \delta)$ in which the leap is less than $\varepsilon$. Eventually, for large enough $n$, the interval $(\xi - \delta, \xi + \delta)$ includes $[a_n, b_n]$, which is a contradiction that $[a_n, b_n]$ has leap greater than $\varepsilon$.

Theorem. Let $f$ be continuous in the closed interval $[a, b]$, then $f$ is uniformly continuous.

Proof.

Suppose $\varepsilon > 0$. Divide $[a, b]$ into a finite number of subintervals in each of which the leap of $f$ is less than $\varepsilon/2$. Take $\delta$ be the length of the smallest subintervals. For any $x_1$ and $x_2$ such that $\vert x_1 - x_2 \vert < \delta$, either they are in the same subinterval or in adjacent ones. In former case, $\vert f(x_1) - f(x_2) \vert < \varepsilon/2$. In the latter case, let $c$ be the common point and $\vert f(x_1) - f(x_2) \vert \le \vert f(x_1) - f(c) \vert + \vert f(c) - f(x_2) \vert < \varepsilon$.

References